Question

A discrete random variable X takes the values 0, 1, 2, 3, 4 and its mean is 1.6. If P(X=1)=0.4, P(X=4)=P(X=2) and P(X=3)=2P(X=2), then P(X=0) is:

• A. 0.2
• B. 0.1
• C. 0.4
• D. 0.3

Answer 1

The Correct Option is A

To solve for \( P(X=0) \), we use the given information about the probabilities of a discrete random variable \( X \) and its mean: \(\mu = 1.6\). The variable \( X \) can take values \( 0, 1, 2, 3, 4 \) with their respective probabilities \( P(X=0), P(X=1), P(X=2), P(X=3), P(X=4) \). The equations presented are:

• The sum of probabilities equals 1:
  \( P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1 \)
• The mean is given as:
  \( E(X) = 0\cdot P(X=0) + 1\cdot P(X=1) + 2\cdot P(X=2) + 3\cdot P(X=3) + 4\cdot P(X=4) = 1.6 \)
• \( P(X=1)=0.4 \)
• \( P(X=4) = P(X=2) \)
• \( P(X=3) = 2 P(X=2) \)

Substitute the known values and relationships into the equations:

• From \( P(X=3) = 2 P(X=2) \), let \( P(X=2) = p \), then \( P(X=3) = 2p \).
• From \( P(X=4) = p \).

Substitute these into the probability sum equation:

• \( P(X=0) + 0.4 + p + 2p + p = 1 \)
• \( P(X=0) + 0.4 + 4p = 1 \)
• \( P(X=0) = 0.6 - 4p \) (Equation 1)

Substitute into the mean equation:

• \( 0.4 + 2p + 6p + 4p = 1.6 \)
• \( 0.4 + 12p = 1.6 \)
• \( 12p = 1.2 \)
• \( p = 0.1 \)

Substitute \( p = 0.1 \) back into Equation 1:

• \( P(X=0) = 0.6 - 4(0.1) \)
• \( P(X=0) = 0.6 - 0.4 \)
• \( P(X=0) = 0.2 \)

Therefore, the probability \( P(X=0) \) is 0.2.